0
votes
0answers
18 views

A filterbase generating filter F

Show that a non empty subset $X$ of a filter $F$ in $B$ is a base for $F$ iff $X$ generates $F$ and for all $x,y$ $\in$ $X$ $\exists$ $z $ $\in$ $X$ such that $z$ $\leqq$ x $\wedge$ y.
3
votes
1answer
151 views

Do filters on a Boolean algebra also make a Boolean algebra?

Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...